---- - Let $X,Y$ be [[topological space]]s; - Let $f: X \to Y$. > [!theorem] Theorem. ([[continuity characterizations]]) > The following are equivalent: > 1. $f$ is [[continuous]]; > 2. (Closure containment) For every subset $A$ of $X$, one has $f(\overline{A})\subset \overline{f(A)}$; > 3. (Closed set characterization) For every [[closed set]] $B \subset Y$, the set $f^{-1}(B)$ is [[closed set|closed in]] $X$; > 4. (Neighborhood nestling) For each $x \in X$ and each [[neighborhood]] $V$ of $f(x)$, there is a [[neighborhood]] $U \ni x$ such that $f(U) \subset V$. > [!proof]- Proof. ([[continuity characterizations]]) > We'll show $(1) \implies (2) \implies (3) \implies (1)$, and then $(1) \iff (4)$. > \ > $(1) \implies (2).$ Assume $f$ is [[continuous]] on $X$ and let $A \subset X$ with $x \in \overline{A}$. We want to show $f(x) \in \overline{f(A)}$. Let $V$ be a [[neighborhood]] of $f(x)$. [[continuous|Continuity]] of $f$ implies that $f^{-1}(V)$ is [[open set|open in]] $X$, after which [[neighborhood-basis characterization of set closure]] further implies $f^{-1}(V) \cap A \neq \emptyset$. Now, applying $f$ to the intersection, [[image of intersection is a subset of intersection of images|we have]]$f(f^{-1}(V) \cap A) \subset V \cap f(A)$ > and since the LHS is nonempty the RHS must not be either. By [[neighborhood-basis characterization of set closure]] we may then conclude $x \in \overline{f(A)}$. > > $(2) \implies (3)$. Assume $(2)$ holds, so that for all $A \subset X$ one has $f(\overline{A}) \subset \overline{f(A)}$. Let $B \subset Y$ be [[closed set|closed in]] $Y$ and set $A := f^{-1}(B)$; it'll suffice to show $x \in \overline{A} \implies x \in A$ since then $\overline{A} \supset \subset A$. Supposing $x \in \overline{A}$, write $f(x) \in f(\overline{A}) \subset \overline{f(A)} = \overline{B} = B,$ > so that $x \in f^{-1}(B)=A$ as desired. > > $(3) \implies (1)$. Assume $(3)$ holds. Let $V \subset Y$ be [[open set|open in]] $Y$ and set $B=Y-V$. Then $X - f^{-1}(V) = f^{-1}(Y-V)=f^{-1}(B)$ > is [[closed set|closed in]] $X$ by $(3)$, hence its complement $f ^{-1}(V)$ is [[open set|open in]] $X$. Therefore $f$ is [[continuous]]. > > $(1) \implies (4)$. Assume $(1)$ holds: $f$ is [[continuous]] on $X$. Let $x \in X$ and $V \subset Y$ a [[neighborhood]] of $f(x)$. Then $f^{-1}(V) =: U$ is a [[neighborhood]] of $x$ since $f$ is [[continuous]]. Observe $ff^{-1}(V) = f(U) \subset V$. > > $(4) \implies (1)$. Let $V$ be [[open set|open in]] $Y$; let $x \in f^{-1}(V)$. By $(4)$ there exists a [[neighborhood]] $U_{x} \ni x$ such that $f(U) \subset V$. We want to show $f^{-1}(V)$ is [[open set|open in]] $X$. This follows from the fact that $f^{-1}(V)$ can be written as the union of such sets $U_{x}$. ---- #### ----- #### References > [!backlink] > ```dataview TABLE rows.file.link as "Further Reading" FROM [[]] FLATTEN file.tags GROUP BY file.tags as Tag > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```